A company manufactures high-powered fountains that can be used to create elabora

Question

A company manufactures high-powered fountains that can be used to create elaborate water shows set to light and music. The company makes three models of fountains, each of which can launch a vertical stream of water to a different height, as indicated in the table below.

Time to fall(s)

Complete the following tasks, filling in the missing values in the table as you find them:

Find the time water would take to fall from each height in the table back to its starting level. Round your answer to the nearest thousandth of a second.

Find the initial upward velocity required for fountain B to launch the water to its maximum height. Round your answer to the nearest tenth of a foot per second.

Create a function that will give you the minimum initial upward velocity, v, required to launch water to any given height,h.

ModelA ModelB ModelC Height(ft) 77 100 240

Time to fall(s)

2.194 Initial Velocity(ft/s) 70.2 123.9

Explanation / Answer

we can use 3 useful equations of motion

1) v = u +at

2) s = ut +(1/2) at2

3)v2 - u2 =2as

where v = final velocity, u = initial velocity, a =acceleration of particle, s= distance covered , t = time taken

So for our queation when water jet is going upward its final velocity will be zero

hence for ModelA

s =h(t) = 77ft , t = 2.194, u = 70.2, v = 0

using 3rd equation of motion

0 - (70.2)2 = 2a *(77)

=> 2a = -(70.2)2/77   = - 64.0005194805        

=> a= -32.0002597403          (it is acceleration due to gravity and will be negative when jet is going upward, +ve when coming donq)

Now for model B) s= h(t) = 100ft

v = 0 in upward motion

3rd eaution of motion will gives

02 - u2 = (2a)* 100

=>     u2 = -(2a)* 100 = (64.0005194805) *100

=> u = 10 * sqrt(64.0005194805)

         = 10* 8.00003246747 ft/sec

          = 80.0003246747   ft/sec

now to get the time to come back to original position, we have s= h(t) = 100ft, u = 0 (initial velocity will be zero in downward motion)

use 2nd eqution of motion

s = ut +(1/2) at2

100 = 0 * t + (1/2) *32.0002597403 t2                       (a is positive in downward motion)

=> 200/32.0002597403 = t2

=> t =2.49998985396 sec;

and velocity after t sec can be fount using 1st equation

v = u+at

v = 0 + 32.0002597403*2.49998985396

v = 80.0003246747

for modelC, use 2nd eation with u = 0 and s =h(t) = 240ft

240 = 0 + (1/2) *32.0002597403 t2

t2 = 280/32.0002597403 = 8.74992897784

=> t= 2.95802788659 sec;

function can be found using 3rd eation of motion with final velocity =0 and initial velocity = v

0 - v2 = (-2a)h                           (acceleration due to gravity will be -ve in upward motion)

v = sqrt(2ah)                      is the answer

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